The column that had yielded before it was loaded

Two lines on the column curve are exact. The squash load is the area times the yield stress and needs no argument; Euler’s load is an eigenvalue and is exact for the problem it states. Between them, over a band of slenderness that contains most of the columns anybody builds, real columns fall below both — and the standard explanation, that they were never straight, accounts for some of the gap and not all of it.

The essay on the column that was never straight makes the other argument, and the two are independent: one is about geometry, this one is about material, and a perfectly straight column made of steel that had cooled unevenly still lands under both bounds.

The stress that arrived with the section #

A rolled I-section leaves the mill at around 1,100 °C and cools in air. The flange tips are exposed on three sides, the web-flange junctions on almost none, so the tips cool first. Cooling steel contracts; steel that is already cold and stiff resists that contraction; and by the time the whole section is at room temperature the parts that cooled first are held in compression and the parts that cooled last in tension. The same process leaves a welded plate girder with a worse pattern still, because a weld bead is the last thing on the section to cool.

The magnitudes are of the order of 0.3 of the yield stress for a rolled section and can reach 0.5 for a welded one, where a weld bead cools last and pulls hard. And the whole field is self-equilibrating: its resultant force is zero, its resultant moment is zero, and no equilibrium equation written about the member can detect it.

That is why it costs no squash load. Load a stub column and the tips yield early, the rest of the section catches up, and at full plasticity every fibre is at $f_y$ regardless of what it started at — so the squash load is exactly $Af_y$ and the residual stress has vanished from the answer. It costs nothing at the strength limit and everything at the stability limit, and the reason is that buckling is not a strength question.

Which free body produced the number #

Cut the column at any station and take the piece above the cut. What crosses it is the axial force $P$ and, once the column starts to bend, a moment. The bending moment is resisted by the section’s stiffness — and stiffness, unlike strength, is supplied only by material that is still on the elastic part of its stress-strain curve.

So the free body’s stiffness is $E \cdot I_e$, where $I_e$ is the second moment of the elastic core — the part of the section that has not yet reached $f_y$. That is the tangent-modulus concept, and it is usually written as an effective modulus $E_t = E \cdot I_e/I$ so that Euler’s expression can be reused:

$$\sigma_{cr} = \frac{\pi^2 E_t}{\lambda^2} = \frac{\pi^2 \tau E}{\lambda^2}, \qquad \tau = \frac{I_e}{I}$$

which is one equation in one unknown, since $\tau$ depends on $\sigma$ and $\sigma$ is what is being solved for. It is the energy criterion with a modulus that has become a function of the answer, and it has to be iterated for that reason. The solver behind these figures finds it by bisection, and the root is unique because $\tau$ falls monotonically as $\sigma$ rises.

The exponent, which is the finding #

With the residual stress varying linearly across the flange from $-\sigma_r$ at the tips to $+\sigma_r$ at the centre, a uniform applied compression $\sigma$ yields the outer part of the width and leaves an elastic core of fraction

$$\xi = \frac{f_y - \sigma}{\sigma_r}$$

Now the two axes diverge, and they diverge because the flange plays a different geometric role in each.

About the major axis, the flanges are lever arms. Their contribution to $I$ is $A d^2/4$ and it is proportional to the area still elastic, so $\tau = \xi$.

About the minor axis, each flange bends about its own centreline. Its contribution is $t b^3/12$, and deleting a strip from each edge leaves a core of width $\xi b$, so $\tau = \xi^3$.

At 80% of yield: the major axis retains 67% of its stiffness and the minor axis 30%. At 90%: 33% and 3.7%. The weak axis loses stiffness as the cube of what the strong axis loses, and the weak axis is the one that governs an unbraced column.

The consequence for the curve is real but more modest than the exponent suggests, and it is worth saying so plainly. At $\lambda = 20$ the minor axis gives $0.884 f_y$ against the major’s 0.980; at $\lambda = 60$, 0.766 against 0.842; at $\lambda = 77$, 0.730 against 0.769. The exponent is doing its work, but the buckling stress is a fourth root of the stiffness times a slenderness, so a factor of two in $\tau$ is a much smaller factor in $\sigma$.

The band it lives in, and the two ends it stops at #

Two slendernesses bound the effect and both are computable.

Above $\lambda_p = \pi\sqrt{E/(f_y - \sigma_r)} = 91$, the Euler stress is below $f_y - \sigma_r$ and nothing has yielded when the column buckles. The elastic answer stands exactly, which the figure shows as the three curves merging.

Below $\lambda_y = \pi\sqrt{E/f_y} = 76$, the Euler line is above the squash line, so the straight answer is $f_y$ — and the tangent-modulus curve is below it and stays below it at every slenderness. A column with residual stress never reaches its squash load at any slenderness, on this model, which is the model’s own weakness as much as its finding: a very stubby column does not buckle at all, it squashes, and the criterion $\sigma = \pi^2\tau E/\lambda^2$ is the wrong question to ask of it.

The honest reading is therefore: the tangent-modulus reduction is genuine and large between about $\lambda = 40$ and $\lambda = 91$, is being extrapolated below 40, and vanishes above 91.

Both of those bounds are built out of $f_y$, which means the band is not a fixed range of slendernesses at all. It slides when the grade changes.

That figure corrects a common intuition, which is that a stronger steel escapes the effect. It does not. The residual stress scales with the yield stress rather than with anything absolute, so the proportional loss is the same at both grades; what moves is which columns suffer it, and it moves down the axis, onto members a designer reading a table of slenderness limits would think of as stocky. A high-grade column is not further from this mechanism than a mild-steel one, it is a different length away from it.

The other end of the argument is the section with no residual stress at all, which is worth drawing because it is the only way to see how much of the sag belongs to this mechanism and how much belongs to the shape of the two straight lines.

So the sag is not an artefact of the plotting and it is not the crossover being drawn as a curve. With $\sigma_r = 0$ there is no sag anywhere; with $\sigma_r = 0.3f_y$ there is a 27 per cent one over a fourteen-unit band of slenderness. Everything between those two figures is one input.

Two theories, and the one that was right for the wrong reason #

Engesser proposed the tangent-modulus load in 1889, was told by Considère that it must be wrong, and replaced it in 1895 with the reduced-modulus load — which recognises that when a column starts to bend, one side unloads elastically while the other continues to yield, so the effective modulus is a weighted mean of $E$ and $E_t$ and lies above $E_t$.

The reduced-modulus load is higher, and it is the theoretically correct bifurcation load for a column that is loaded first and bent afterwards. Tests came in below it and near the tangent-modulus load, and the discrepancy stood for fifty years.

Shanley resolved it in 1947 with an argument that is one of the most satisfying in the subject. The two theories answer different questions. The reduced-modulus load assumes the column stays straight until it buckles, so that unloading can occur. But there is nothing to stop a column bending while the load is still increasing — and if it does, no fibre ever unloads, so the tangent modulus applies. The tangent-modulus load is therefore the load at which bending can begin, and the reduced-modulus load an upper bound on where it ends up.

A real column starts to bend at the tangent-modulus load, stiffens slightly as it does, and reaches a maximum somewhere between the two — much nearer the lower one. The lower bound was right because it was answering the question a column actually asks.

What the codes do instead, and why they look different #

There is a second mechanism that produces a sag in the same place, and a code fits both at once without separating them. An initial bow is amplified by $1/(1 - P/P_{cr})$, grows without limit as the load approaches the critical value, and produces a moment the section carries alongside its axial force — so first yield of a bent member arrives well before the Euler load does. That is a statement about geometry and it uses none of the machinery on this page.

Design codes do not compute a tangent modulus. They fit a Perry-type curve — an equivalent initial bow, chosen so that first yield of the imperfect member reproduces the test data — and then publish several such curves labelled by the section’s manufacture and its axis of buckling.

That labelling is this page’s argument in disguise. A rolled H-section buckling about its major axis is assigned a higher curve than the same section buckling about its minor axis, and the two differ by nothing geometric that the imperfection model can see. The difference is $\xi$ against $\xi^3$, absorbed into a fitted imperfection factor because the fitted form is easier to tabulate.

So the equivalent imperfection in a code is not an imperfection. It is a bookkeeping device carrying at least three physical effects — geometric bow, residual stress, and the variation of yield stress through the section — and the fact that a welded box gets a different curve from a rolled H is evidence that the second of those is doing much of the work.

That is the separation between two of a code’s buckling curves, arrived at from the mechanism rather than from a table. A fitted imperfection can reproduce the gap between those two curves and it cannot explain why the gap is there, why it is wider about one axis than the other, or why it extends further up the slenderness range for the welded member. All three fall out of one number being 0.5 rather than 0.3.

The same argument one scale down, and one scale up #

The mechanism generalises in both directions, and in both directions it is the same sentence: a yielded fibre carries force and supplies no stiffness.

Downward, to the plate: a rippling flange is a stability problem in a strip of the same section, and the residual compression that yields the flange tips is at the plate’s unstiffened edge, which is where its buckling stress is set. The two calculations are usually done separately and they are about the same steel in the same place.

Upward, to the frame: the same softening feeds the restraint each column offers the next and the sway stiffness of the storey. A column at 80% of yield about its minor axis has 30% of its stiffness, and a beam-to-column joint restrained by such a member is restrained by very much less than the drawing suggests. Inelasticity is not local to the member it happens in, which is the part of the argument no member-by-member check contains.

And sideways, along the member itself. Nothing in a tangent-modulus calculation is uniform along a column’s length. A column that has begun to bend carries its largest moment at mid-height, yields there first, and is therefore softer in the middle than at its ends — so the honest object is not a column with a reduced modulus but a column with a reduced modulus over part of its length.

Thirty-six per cent on the unsafe side is a large error for an operation as innocent as taking a mean, and the reason it happens is the reason this page is about stiffness rather than strength. A critical load is set by where the curvature is, not by how much material there is; the third of the column being averaged over is the third where the mode has all of its curvature, so softening it does more than its share of the damage. A column yielding at mid-height has been softened in exactly the segment that decides its answer.

The same statement one scale smaller is the section’s own moment-curvature relation, which is a straight line only while every fibre is elastic and flattens past first yield. That flattening is the loss this page has been computing: a column’s tangent modulus and a section’s moment-curvature slope are one quantity read at two scales.

The same yielding costs the two axes different amounts #

The residual compression sits at the flange tips, which is where the pattern puts it and where the metal cooled first. That location decides something the essay has so far treated as one number, because the tips are not equally important to the two axes.

Suppose the outer 15 per cent of each flange has yielded, so 30 per cent of the flange area is inert and an elastic core of 70 per cent of the width survives.

About the major axis, the flanges contribute through $A_f d^2/4$ — every part of them is at nearly the same distance from the axis — so losing 30 per cent of the area loses 30 per cent of the stiffness. The effective $EI_x$ is 0.70 of the elastic value.

About the minor axis, the flanges contribute through $t b^3/12$, and the surviving core is 0.7 of the width:

$$\frac{E_tI_y}{EI_y} = 0.7^3 = 0.343$$

Thirty-four per cent. The identical yielded region costs the minor axis twice what it costs the major one, because the material that yielded first is the material furthest from the minor axis and therefore the material that axis depends on most.

Which is the reason a section is assigned different buckling curves for its two axes — a fact that looks like a calibration in a code table and is a geometric consequence of where the residual compression lives. It also compounds a disadvantage the section already had: the minor axis was the weak one to begin with, it is the axis a column usually buckles about, and it is the one the residual pattern damages hardest.

The same argument says what a welded section does. Welding puts yield-level compression at the flange tips too, and rather more of it, so the effect is larger by the same route — which is why welded sections sit on a lower curve than rolled ones of the same shape, and why the gap between the two curves is wider about the minor axis than about the major.

Six point four per cent against twenty-nine is the cube law working on a smaller core, and it is a much larger difference than the change in the input. Twenty percentage points more residual stress removes four-fifths of what was left of the weak-axis stiffness at that load, because the exponent is applied to a core that the extra residual stress has already narrowed.

It suggests the one remedy available, too, and it is a shape rather than a treatment. Move the compression away from the tips and the damage falls: a hollow section has no tips, its residual pattern is far more uniform round the perimeter, and every part of it is at a similar distance from both axes. That is one of the reasons a square or circular hollow section sits on the highest of the buckling curves — not only because it has no weak axis, but because whatever residual stress it carries is distributed where it costs least.

Where the model stops #

The residual pattern is idealised as a straight line across the flange. Measured patterns are curved, differ between flange and web, differ between rolled, welded and flame-cut sections, and vary along a member’s own length. The 0.3 used here is a representative peak value, not a property of steel.

The core is assumed to stay symmetric. With a symmetric residual pattern and a concentric load it does; with a load applied even slightly off centre it does not, the yielded region is one-sided, and the section’s centroid of stiffness moves — which introduces a bending moment out of nothing, in the way an unsymmetric section bends about an axis nobody drew.

Nothing here is an imperfection. The whole page assumes a perfectly straight column and a perfectly concentric load, which is why it is a separate argument from the previous one. A real column has both effects at once, and they are not additive: the bow produces a moment, the moment yields one flange preferentially, and the yielding softens the very stiffness the bow’s amplification depends on.

Below about $\lambda = 40$ the model is being asked the wrong question, as set out above. It says a stub column reaches 88% of squash; a stub column reaches squash.

And the tangent-modulus argument assumes a plateau. The whole of the stiffness loss depends on a yielded fibre supplying nothing, which is true of mild steel and its long flat plateau. For a material with a rounded curve and no plateau — aluminium, stainless steel, high-strength steel with an 0.2% proof stress — $E_t$ falls gradually from the proportional limit, there is no sharp core, and the column curve has to be computed from the material’s own tangent modulus rather than from a geometry of yielded strips.

What the pictures cannot show #

The core figure draws a sharp boundary between yielded and elastic material, which the section does not have. Yield spreads through a real flange as a diffuse front, and the sharp line is a consequence of the elastic-perfectly-plastic idealisation rather than an observation.

Neither of the column-curve figures can show what a reader most wants: the scatter. Column tests do not fall on a curve — they fall in a band 20% wide, because residual stress varies from mill to mill and heat to heat, and because the initial bow varies from member to member. Every curve drawn here is a single realisation of a distribution, and the codes’ curves are lower fractiles of measurements rather than predictions.

And the shading in the core figure implies that the yielded strips are inert. They are not: they carry axial force, at $f_y$, and they carry the largest share of it. What they do not carry is any change of force with a change of curvature, which is a distinction no static picture can draw.

The ladder from here #

Later rungs on this anchor: the tangent modulus of a rounded stress-strain curve, and the Ramberg-Osgood parameters that make an aluminium column curve a different shape rather than a shifted one. Shanley’s model column — two flanges and a hinge — set out in full, since it settles the tangent-versus-reduced question in four lines of algebra. The beam-column, where axial force and moment yield the section together and the interaction is not a curve anybody can write. Column curves as they were actually made: the European buckling curves came from about a thousand tests and a hundred simulations, and reading them as fitted data rather than as theory changes what they mean. The effect of the same residual stress on lateral-torsional buckling, where it lowers the elastic critical moment by the same mechanism. And the welded box column, where the residual pattern is worse, the tips are the corners, and the two axes are the same.

The historical part is the piece worth carrying. For half a century the profession used a formula it had been told was theoretically wrong, because it agreed with the tests; and when the theory was finally repaired, the repair showed that the formula had been answering a better-posed question all along.